import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;

public class BinaryTree {
    class TreeNode {
        public char val;
        public TreeNode left;
        public TreeNode right;
        public TreeNode(char val) {
            this.val = val;
        }
    }

    // 获取树中节点的个数
    public TreeNode createTree(){
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');

        TreeNode root = A;
        root.left = B;
        root.right = C;
        B.left = D;
        C.left = E;
        C.right = F;
        return root;
    }

    //前序遍历
    public void preOrder(TreeNode root){
        if(root == null){
            return;
        }
        System.out.print(root.val + " ");
        preOrder(root.left);
        preOrder(root.right);
    }

    //中序遍历
    public void inOrder(TreeNode root){
        if(root == null){
            return;
        }
        inOrder(root.left);
        System.out.print(root.val + " ");
        inOrder(root.right);
    }

    //后序遍历
    public void postOrder(TreeNode root){
        if(root == null){
            return;
        }
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val + " ");
    }

    // 获取树中节点的个数
    public int size(TreeNode root){
        if(root == null){
            return 0;
        }
        return size(root.left) + size(root.right) + 1;
    }

    // 获取叶子节点的个数//子问题
    public int getLeafNodeCount(TreeNode root){
        if(root == null){
            return 0;
        }
        if(root.left == null && root.right == null){
            return 1;
        }
        return getLeafNodeCount(root.left) + getLeafNodeCount(root.right);
    }

    // 获取第K层节点的个数
    public int getKLevelNodeCount(TreeNode root,int k){
        if(root == null || k <= 0){
            return 0;
        }
        if(k == 1){
            return 1;
        }
        return getKLevelNodeCount(root.left,k-1) + getKLevelNodeCount(root.right,k-1);
    }

    // 获取二叉树的高度  时间复杂度：O(n)
    public int getHeight(TreeNode root) {
        if(root == null){
            return 0;
        }
        int leftHigh = getHeight(root.left);
        int rightHigh = getHeight(root.right);

        return leftHigh > rightHigh ? leftHigh + 1 : rightHigh + 1;
    }

    // 检测值为value的元素是否存在
    public TreeNode find(TreeNode root, char val) {
        if(root == null){
            return null;
        }
        if(root.val == val){
            return root;
        }

        TreeNode leftTreeRet = find(root.left,val);
        if(leftTreeRet != null){
            return leftTreeRet;
        }

        TreeNode rightTreeRet = find(root.right,val);
        if(rightTreeRet != null){
            return rightTreeRet;
        }

        return null;
    }

    //层序遍历
    public void levelOrder(TreeNode root){
        if(root == null){
            return;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            TreeNode cur = queue.poll();
            System.out.print(cur.val + " ");
            if(cur.left != null){
                queue.offer(cur.left);
            }
            if(cur.right != null) {
                queue.offer(cur.right);
            }
        }
    }

    // 判断一棵树是不是完全二叉树
    public boolean isCompleteTree(TreeNode root) {
       if(root == null){
           return true;
       }
       Queue<TreeNode> queue = new LinkedList<>();
       queue.offer(root);
       while(!queue.isEmpty()){
           TreeNode cur = queue.poll();
           if(cur != null){
               queue.offer(cur.left);
               queue.offer(cur.right);
           }else{
               break;
           }
       }
       while(!queue.isEmpty()){
           if(queue.poll() != null){
               return false;
           }
       }
       return true;
    }

}
